Question: Divide the following rational expressions and simplify the result. $\dfrac{x^4-625}{25+10x+x^2} \div \dfrac{7x^2-37x+10}{-8x-40}=$
Explanation: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $x^4-625$, of the first expression can be factored as $(x^2+25)(x^2-25)$ using the difference of squares pattern. Then, we can further factor it to $(x^2+25)(x+5)(x-5)$ by using the difference of squares pattern again. The denominator, $25+10x+x^2$, of the first expression can be factored as $(x+5)(x+5)$ using the perfect squares pattern. The numerator, $7x^2-37x+10$, of the second expression can be factor by grouping as $(x-5)(7x-2)$. The denominator, $-8x-40$, of the second expression can be factored to $-8(x+5)$ by factoring out $-8$. Now the quotient looks as follows: $\dfrac{(x^2+25)(x+5)(x-5)}{(x+5)(x+5)} \div \dfrac{(x-5)(7x-2)}{-8(x+5)}$ To find the quotient of two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\phantom{=}\dfrac{(x^2+25)(x+5)(x-5)}{(x+5)(x+5)} \div \dfrac{(x-5)(7x-2)}{-8(x+5)}$ $\begin{aligned} &= \dfrac{(x^2\!+\!25)(x\!+\!5)(x\!-\!5)}{(x\!+\!5)(x\!+\!5)} \cdot \dfrac{-8(x\!+\!5)}{(x\!-\!5)(7x\!-\!2)}&\text{Flip the divisor.}\\\\\\ &= \dfrac{(x^2\!+25)(x+5)(x-5) \cdot -8(x+5)}{(x+5)(x+5) \cdot (x-5)(7x-2)} &\text{Multiply across.}\\\\\\ &= \dfrac{(x^2\!+\!25)\!{\cancel{(x\!+\!5)}}\!{\cancel{(x\!-\!5)}}\!( -8)\!{\cancel{(x\!+\!5)}}}{{\cancel{(x\!+\!5)}}\!{\cancel{(x\!+\!5)}} \!{\cancel{(x\!-\!5)}}\!(7x\!-\!2)} &\text{Cancel common factors.}\\\\\\\\ &=\dfrac{-8(x^2+25)}{7x-2} \end{aligned}$ Therefore, the simplified form of the quotient is $\dfrac{-8(x^2+25)}{7x-2}$, which is equivalent to $\dfrac{-8x^2-200}{7x-2}$.